A059329 Number of 3 X 3 matrices, with elements from {0,...,n}, having the property that the middle element of each of the eight 3-element horizontal, vertical and diagonal lines equals the average of the two end elements.

1, 2, 7, 12, 25, 38, 63, 88, 129, 170, 231, 292, 377, 462, 575, 688, 833, 978, 1159, 1340, 1561, 1782, 2047, 2312, 2625, 2938, 3303, 3668, 4089, 4510, 4991, 5472, 6017, 6562, 7175, 7788, 8473, 9158, 9919, 10680, 11521, 12362, 13287, 14212

Offset: 0

Author

John W. Layman, Jan 26 2001

Keywords

• nonn
• easy

Comments

The bisections of the first differences of {a(n)} give A001844 (the centered triangular numbers n^2+(n-1)^2).

Quasipolynomial of order 2. - Charles R Greathouse IV, Dec 07 2011

Also, the number of 3 X 3 magic squares with elements from {0,...,n} and with duplicate elements allowed. If [[a,b,c], [d,e,f], [g,h,i]] satisfies the property in the description of this sequence, then [[h,a,f], [c,e,g], [d,i,b]] is a magic square, and conversely. - David Radcliffe, Apr 13 2020

Links

• G. C. Greubel, Table of n, a(n) for n = 0..1000
• Ronald Cools, Ian H. Sloan, Minimial cubature formulae of trigonometric degree, Math. Comp. 65 (216) (1996) 1583-1600. Table 1 dimension 3.
• Index entries for linear recurrences with constant coefficients, signature (2,1,-4,1,2,-1).

Programs

Mathematica

LinearRecurrence[{2, 1, -4, 1, 2, -1}, {1, 2, 7, 12, 25, 38}, 50] (* G. C. Greubel, Jan 07 2017 *)

PARI

a(n)=if(n%2,4*n^3+2*n,4*n^3+6*n^2+8*n+3)/3 \\ Charles R Greathouse IV, Dec 07 2011

Formula

From Frank Ellermann: (Start)

even: a(2*n) = (4*n^3 + 6*n^2 + 8*n + 3)/3.

odd: a(2*n-1) = (4*n^3 + 2*n)/3. (End)

a(n) = Sum_{k=0..n} A109613(k)*A109613(n-k). - Reinhard Zumkeller, Dec 05 2009

From Colin Barker, Mar 29 2013: (Start)

a(n) = ((1+n)*(9+3*(-1)^n+4*n+2*n^2))/12.

G.f.: (x^2+1)^2 / ((x-1)^4*(x+1)^2). (End)

E.g.f.: (1/12)*(3*(1 - x)*exp(-x) + (9 + 21*x + 12*x^2 + 2*x^3)*exp(x)). - G. C. Greubel, Jan 07 2017